Hooke's law

Further Maths · Further Mechanics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

An elastic string or spring has a natural length, its length when no force acts on it. Hooke's law says the tension is proportional to the extension, which is the stretched length minus the natural length. The modulus of elasticity λ is a force in newtons: it is the tension that would double the natural length.

A string can only pull, and it goes slack when it is shorter than its natural length. A spring can also be compressed, and then it pushes with a thrust given by the same formula.

A stretched or compressed string or spring stores elastic potential energy. For motion problems, use conservation of energy with kinetic, gravitational and elastic energy. At an equilibrium position, balance the forces instead.

Rules to remember

Hooke's law: tension = modulus × extension ÷ natural length (T = λx/l)
Elastic potential energy = modulus × extension² ÷ (2 × natural length) (EPE = λx²/(2l)), in joules; this also equals ½Tx
Extension = stretched length − natural length; never put the full length into the formulae
Particle hanging at rest on a vertical string: λe/l = mg, where e is the extension at equilibrium
Energy: kinetic energy + gravitational potential energy + elastic potential energy is constant when no other force does work; subtract the work done against friction if the surface is rough
Speed is greatest at the equilibrium position (acceleration zero); the particle is at rest for an instant at the greatest extension
A slack string has zero tension and zero elastic potential energy

Common mistake

Students find the greatest extension by balancing forces (T = mg). That gives the equilibrium position, where the speed is greatest. The lowest point is where the speed is zero, and it must be found with an energy equation.

Worked example

A particle of mass 0.5 kg is attached to one end of a light elastic string of natural length 1 m and modulus of elasticity 20 N. The other end is fixed at O. The particle is released from rest at O. Take g = 10 m s⁻². Find the greatest distance of the particle below O.

Let the greatest extension be x m. The particle is at rest at the start and at the lowest point, so loss of gravitational potential energy = gain of elastic potential energy.
0.5 × 10 × (1 + x) = 20x² ÷ (2 × 1), so 5 + 5x = 10x².
2x² − x − 1 = 0, so (2x + 1)(x − 1) = 0 and x = 1 (the extension must be positive).
Greatest distance below O = 1 + 1 = 2 m.

Answer: 2 m

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